Answer : The percent by volume of oxygen gas in the mixture is, 5.3 %
Explanation :
According to the Raoult's law,
![p^o_A=X_A\times p_A](https://tex.z-dn.net/?f=p%5Eo_A%3DX_A%5Ctimes%20p_A)
where,
= total partial pressure of solution = 3.8 atm
= partial pressure of oxygen = 0.20 atm
= mole fraction of oxygen = ?
Now put all the given values in this formula, we get:
![p^o_A=X_A\times p_A](https://tex.z-dn.net/?f=p%5Eo_A%3DX_A%5Ctimes%20p_A)
![0.20atm=X_A\times 3.8atm](https://tex.z-dn.net/?f=0.20atm%3DX_A%5Ctimes%203.8atm)
![X_A=0.0526](https://tex.z-dn.net/?f=X_A%3D0.0526)
Now we have to calculate the percent by volume of oxygen gas in the mixture.
The mole percent of oxygen gas = ![0.0526\times 100=5.3\%](https://tex.z-dn.net/?f=0.0526%5Ctimes%20100%3D5.3%5C%25)
As we know that, there is a direct relation between the volume of moles.
So, mole percent of oxygen gas = volume percent of oxygen gas
Volume percent of oxygen gas = ![0.0526\times 100=5.3\%](https://tex.z-dn.net/?f=0.0526%5Ctimes%20100%3D5.3%5C%25)
Therefore, the percent by volume of oxygen gas in the mixture is, 5.3 %